I am using Talend Studio for Data Integration v5.3.1.
In that I created a Job for fileDelimited. I have uploaded a CSV file and it is reading the file.
I exported the Job as a Zip file, extracted it, and I run the sh file in Terminal. And it was reading the file and displaying it in the console.
Now I want to read a different file in some other file location. is it possible to read the different file by running the same shell script? If so, where I have to change?
you can do it using context variables, and context load.
create configuration file which will have all the required input location path and other details.
you have different files on different location and you just wanted upload files without file parsing right? if so then
first create configuration file with two parameters.
FilePath|FileName
\\Folderone\|File1.txt
\\Foldertwo\|File2.txt
create two context variables in context named as FilePath & FileName
and then used tContextLoad to load above configuration details to context variable
used these variables to provide file and path at runtime from configuration.
See my answer here : https://stackoverflow.com/a/17565718/921244 to have guidance on how to open the Demo project.
There you will find a demo job for loading Context variables.
If you want an online example, take a look at the official documentation : http://www.talendforge.org/tutorials/tutorial.php?language=english&idTuto=34
Related
I am currently working on a project using jmeter-maven plugin. I need to use a CSV data config file from which some variables are to be read during execution. Although the created JMX file works in the JMeter GUI, the same isn't working in non-gui mode.
From the logs, I was able to gather that the data file isn't being opened (stored) during execution in the non-gui mode, which on the other hand, it happens in the GUI mode leading to successful execution in GUI mode.
I have checked the path of the file (the absolute path of the file - with forward slashes), permissions of the file and all the parameters set in the CSV data config element in the jmx file created (it has proper path to the csv file) and I couldn't possibly get the reason on why the CSV file is not being used during execution in non-gui mode.
I have tried having the CSV file in the bin folder (giving the complete path of to the bin folder and just the file name - both methods), having the csv in the same path as the jmx file.
Any ideas on what I might be missing?
CSV Config Data CSV configuration Image
I have a Scala project that I want to export as a jar file so I can run it on another machine(I know how to export as a jar file). My main function reads in an existing input.json file and writes to a output.json file. The path where the the input.json is hard coded in my main and the path where the output.json will be written is hard coded as well. My goal is to export a jar file and pass in multiple input.json files to this jar file and my desire is the output.json file to be written to some directory. Basically, I want to have a large pipeline that feeds many different input.json files to this jar file and outputs all the output.json to some directory. I'm not too sure if this can be done, and if so how exactly?
Try to use the path from the configuration and from the environment variables and if the environment variable is not present it will pick up the path from the Configuration.
Change the environmental variable according to your needs and there you go.!
But in the case you want a whole directory to be taken as input containing multiple json files. Then in that case you have to implement a hack !
Read the directory path from the configuration file or environmental variable, and read it as directory, extract the list of names and perform operations on them inside a map !
I hope I answered your query!
I'm new to Cloudbees. I just opened an account and trying to upload a .JAR file which basically downloads a file to the location mentioned by user (through java command line arguments). So far I have run the .JAR in my local. So far, I was referring to my local file system to save the file. If I deploy my .JAR file to Cloudbees SDK, where can I save the downloaded file (and then process it).
NOTE: I know this is not a new requirement in java if we deploy the jar in UNIX/WINDOWS OS where we can refer the file system w.r.t to home directory.
Edit#1:
I've seen couple of discussions about the same topic.
link#1
link#2
Everywhere they are talking about the ephemeral (temporary) file system which we can access through System.getProperty("java.io.tempDir"). But I'm getting null when I access java.io.tempDir. Did anyone manage to save files temporarily using this tempDir?
You can upload a jar with the java stack specifying the class and classpath (http://developer.cloudbees.com/bin/view/RUN/Java+Container)
Our filesystem however is not persistent, so if you are talking about saving a file from within your application, you could save it in this path
System.getProperty("java.io.tmpdir")
but it will be gone when your application hibernates, scales-up/down or is redeployed.
If you want a persistent way to store file/images, you can use AmazonS3 from your cloudbees application: uploading your files there will ensure their persistence.
You can find an example of how to do that in this clickstart:
http://developer-blog.cloudbees.com/search?q=amazon+s3
More information here: https://wiki.cloudbees.com/bin/view/RUN/File+system+access
Android uses a static resource file R. This file (at least in eclipse) is automatically updated when ever you add new id's of any sort. How can I create/implement the same feature in a normal java application? Is it as simple as just writing an xml parser and just updating the resource file after the xml is modified?
In a way, yes. You need to create a custom build script/program which runs at the start of each build (before anything else), scans your resource folder files (and if they are XML files it needs to read in the XML files and parse out the string resources or whatever from those), then write it all to a Java file in some manner (e.g. R.string_name = "string value").
Make sure the XML files aren't actually packaged in your .jar, since all that information will be stored inside your Java resources file now.
For things which aren't XML files you could just store the filename as a string in the Java resources file.
You didn't specified the type or the use of the resources. I don't know android, but I'll try to help; If you just need to access some resource in your application you can use properties or resource, there are some differences see this other question Properties vs Resource Bundle
Anyone plz let us know what to do when we have some configuration file which is basically xml.I want to for example give the path to save the image(for my java program) in a folder from some config file (xml in my case).In that case where should the config file be kept.Rt now every thing is converted to jar file when i create a java standalone package.But i want to give some setting from xml file.What to do in that case.How is it possible.This article only provides to create a single jar file for java project but talks nothing about the configuration settings that u can provide from some external source.
Regards
Sagar
I'm not sure I fully understand your question, but if it is where to put the XML file with configuration information, you can place your xml file in the same directory as your jar file, and then pass the XML file name and path into the Jar on the command line when calling the Jar. If you're running it in Windows, this is often done using a shortcut. Then you can get the full path string for the Jar from the main method's String[] arg array that accepts the command parameters.
Sagar,
The fact your java program is a standalone package (.jar file) has no bearing on where your configuration file is stored. Your java package is a program and that program can read any file from the file system that it so desires; it does not have to be part of the code inside the IDE i.e. you don't have to write it when you write the program. What you do need is some way, when you start the program, to find and read said configuration file.
Depending on how you expect the program to be configured, you might put that file in a number of locations. For example, /etc/yourimageprogram/config.xml or c:\program files\yourimageprogram\config.xml or perhaps c:\users\Sagar\Application Settings\yourimageprogram\config.xml. Which you choose of those options really depends on what the use case is and that I can't help with.
However, there are some main points to reading any file:
Does it exist?
Are we allowed to open it for reading?
Are we allowed to open it for writing? Might want to know if we want to update the config?
In Java, typically, you would test this with:
File configfile = new File("C:\test.xml");
if ( configfile.exists() && configfile.canRead() )
{
// read the file
}
else
{
// decide what to do if no config exists.
// might be first run of app.
}
The next stage is to parse the file. There are a number of parsers available for XML including sax and org.w3c.dom. What you need to do is to use these to extract the information you require and store that in a class. Probably a singleton class as you're unlikely to have multiple configuration instances per instance of the program.
I suggest you read about XML Parsers and File Handling under Java. Also look at the File object. See all your options for file io in java. These should give you some indication of how to proceed.